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Q6E

Expert-verifiedFound in: Page 433

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In problems 1-6, determine the convergence set of the given power series.**

$\mathbf{\sum}_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty}}\frac{\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{!}}{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{)}}^{{\mathbf{n}}}$

** **The set is, $x\in (-3,-1)$

Use the ratio test to determine the radius of convergence (with ):${a}_{n}=\frac{(n+2)!}{n!}$

$\begin{array}{c}\underset{n\to \infty}{\mathrm{lim}}\left|\frac{{a}_{n}}{{a}_{n+1}}\right|=\underset{n\to \infty}{\mathrm{lim}}\left|\frac{\frac{(n-2)!}{n!}}{\frac{(n+3)!}{(n+1)!}}\right|\\ =\underset{n\to \infty}{\mathrm{lim}}\left|\frac{(n+2)!}{(n+3)!}\cdot \frac{(n+1)!}{n!}\right|\\ =\underset{n\to \infty}{\mathrm{lim}}\left|\frac{(n+2)!}{(n+3)(n+2)!}\times \frac{(n+1)n!}{n!}\right|\\ =\underset{n\to \infty}{\mathrm{lim}}\left|\frac{(n+1)}{n+3}\right|\\ =\underset{n\to \infty}{\mathrm{lim}}\left|\frac{(1+(1/n\left)\right)}{1+(3/n)}\right|\\ =1\end{array}$

The radius of convergence is 1, therefore convergent set for the given power series is.$|x+2|<1$

To completely identify the convergence set, we have to check whether the boundary points -1 and -3 are included in the set or not.

Checking at$x=-1$ , by substituting the value ofx by -1 .

$\begin{array}{c}\sum _{n=0}^{\infty}\frac{(n+2)!}{n!}{\left(x+2\right)}^{n}=\sum _{n=0}^{\infty}\frac{(n+2)!}{n!}{\left(-1+2\right)}^{n}\\ =\sum _{n=0}^{\infty}\frac{(n+2)!}{n!}{\left(1\right)}^{n}\\ =\sum _{n=0}^{\infty}\frac{(n+2)(n+1)n!}{n!}\\ =\sum _{n=0}^{\infty}(n+2)(n+1)\\ =\infty \end{array}$

The above series is divergent, thus the point -1 is excluded from the convergent set

Similarly, checking at $x=-3$ by substituting the value of x by -3.

$\begin{array}{c}\sum _{n=0}^{\infty}\frac{(n+2)!}{n!}{\left(x+2\right)}^{n}=\sum _{n=0}^{\infty}\frac{(n+2)!}{n!}{\left(-3+2\right)}^{n}\\ =\sum _{n=0}^{\infty}\frac{(n+2)!}{n!}\\ =\sum _{n=0}^{\infty}\frac{(n+2)(n+1)n!}{n!}\\ =\sum _{n=0}^{\infty}{(-1)}^{n}(n+2)(n+1)\end{array}$

The above series is divergent, thus the point -3 is not included in the convergent set. The convergent set for the given power series is.$x\in (-3,-1)$

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